Prove Proposition (4.32).
Proposition (4.32) states that
$$ p \text{ prime } \iff \sigma (p) = p + 1 $$
We need to prove both directions:
- $ p \text{ prime } \implies \sigma (p) = p + 1 $
- $ p \text{ prime } \impliedby \sigma (p) = p + 1 $
($\implies$)
We assume $p$ is prime. That means its only divisors are $p$ and 1. The $\sigma$ function is the sum of divisors, which here is $\sigma(p) = p + 1$.
($\impliedby$)
We start with $\sigma(p)=p+1$. This tells us the divisors of $p$ sum to $p+1$. We know the divisors of $p$ include $p$ and $1$, and these sums to $p+1$. This is the value of $\sigma(p)$ and so there is no possibility for additional divisors. That is, the only divisors of $p$ are $p$ and $1$, and so, by definition, $p$ is prime.